!---Just for personal reference
Diagram shows a straight line IJK, a circle, 2 congruent quadrants BEF and CGH, and straight lines extended from the ends of each arc.
The top centre circle is tangent to the upper line at J and the two quadrants at the marked points.
AB and CD are collinear
Given:
p=Distance of FG
q=Distance of BC=Distance if EH
r=Radius of each quadrant
s=Perpendicular distance from IJK to AB or CD
R= Radius of the circle
The relation between p,q,r,s,andR:
(s+r−R)2+(2p+r)2=(R+r)2
Using Pythagorean Theorem
Note :The figure is symmetrical about the vertical diameter line of the circle (ignoring extra lines and points)
#Geometry
Easy Math Editor
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